Find Mass of Passenger Train in Wagon Collision

AI Thread Summary
In a discussion about calculating the mass of a passenger train involved in a collision with a wagon train, the initial problem presented was the loss of 27% of kinetic energy after the collision. Participants noted that the final velocity can be derived from the kinetic energy formula, and emphasized the need for the initial speed of the wagon train, which was not provided. It was clarified that the two trains stick together post-collision, allowing for the application of inelastic collision equations. The conversation highlighted the challenge of having multiple unknowns and the importance of using the given mass of the wagon train to simplify the calculations. Ultimately, the group concluded that with the right approach, the problem could be solved by isolating one unknown variable.
coltsamuel96
Messages
15
Reaction score
0
hello, can someone help me with this,
a wagon train of given mass collides with a passenger train at rest... the train loses 27% of kinetic energy after the collision... find the mass of the passenger train...
so what i did is, i put this formula for inelastic collision,v(final)= m1v1+m2v2/m1+m2... since v2 is in rest , m2=(m1v1/v(final))-m1... so my question is what will be the velocities before and after collision...i did, m2=(m1*1/.73)-m1...but i have a feeling that this is wrong... please help...
 
Physics news on Phys.org
Welcome to PF, coltSamuel.

You can write ½ mv² = .73*½ mV² and find that the reduced velocity is a number multiplied by the original velocity.

I don't see how you can solve for the mass with unknown initial V, unknown final v for the second train and two unknown masses. One equation, 4 unknowns. Hopeless?
Do we know anything else? Do the two trains stick together after the collision?

It is always a good idea to write the question here exactly as it was given to you - maybe one of us can spot some information in there that you missed.
 
the mass of the wagon train is 3.18*10^4kg... and the trains stick together after the collision...
 
You need the initial speed of the train. What is it?
 
there´s no intial velocity... only thing given is the loss of K.E... (27%)
 
Okay, I think we have enough to do it now!
v(final)= m1v1+m2v2/m1+m2
is a good start. One of those v's must be zero since the passenger train was initially at rest. The final v is known to be a certain fraction of the initial velocity (I suggested you work it out with the KE formula earlier). I think you will then be able to cancel the remaining velocity, put in your known mass and have only one unknown remaining!
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top